Difference between revisions of "2020 AIME II Problems/Problem 5"
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<cmath>\equiv 2\cdot24^2 - 1 \equiv 151 \pmod{1000}.</cmath> | <cmath>\equiv 2\cdot24^2 - 1 \equiv 151 \pmod{1000}.</cmath> | ||
− | ==Video | + | ==Video Solutions== |
− | https://youtu.be/lTyiRQTtIZI | + | https://youtu.be/lTyiRQTtIZI |
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+ | https://youtu.be/ZWe_99091e4 | ||
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+ | https://youtu.be/ZhAZ1oPe5Ds?t=5032 | ||
==Video Solution 2== | ==Video Solution 2== |
Revision as of 15:37, 14 February 2021
Contents
[hide]Problem
For each positive integer , let be the sum of the digits in the base-four representation of and let be the sum of the digits in the base-eight representation of . For example, , and . Let be the least value of such that the base-sixteen representation of cannot be expressed using only the digits through . Find the remainder when is divided by .
Solution
Let's work backwards. The minimum base-sixteen representation of that cannot be expressed using only the digits through is , which is equal to in base 10. Thus, the sum of the digits of the base-eight representation of the sum of the digits of is . The minimum value for which this is achieved is . We have that . Thus, the sum of the digits of the base-four representation of is . The minimum value for which this is achieved is . We just need this value in base 10 modulo 1000. We get . Taking this value modulo , we get the final answer of . (If you are having trouble with this step, note that ) ~ TopNotchMath
Solution 2 (Official MAA)
First note that if is the least positive integer whose digit sum, in some fixed base , is , then is a strictly increasing function. This together with the fact that shows that is the least positive integer whose base-eight digit sum is 10. Thus , and is the least positive integer whose base-four digit sum is Therefore
Video Solutions
https://youtu.be/ZhAZ1oPe5Ds?t=5032
Video Solution 2
~IceMatrix
Video Solution 3
https://youtu.be/ZhAZ1oPe5Ds?t=5032
~ pi_is_3.14
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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